Question 54

In $$\triangle$$ABC,P is a point on BC such that BP : PC = 1 : 2 and Q is the mid point of BP. Then, ar($$\triangle$$ABQ): ar($$\triangle$$ABC) is equal to:

Solution

$$\triangle$$ABP and $$\triangle$$APC are having same height and their bases are in ratio 1:2.

So the ratio of their areas would be x:2x

$$\triangle$$ABQ = $$\frac{1}{2}$$ $$\triangle$$ABP = $$\frac{x}{2}$$

ar($$\triangle$$ABQ): ar($$\triangle$$ABC) = $$\frac{x}{2}$$ / 3x = $$\frac{1}{6}$$

So , tha answer would be Option a)1:6 .

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